How do you solve Divide an Array Into Subarrays With Minimum Cost I on LeetCode?

Divide an Array Into Subarrays With Minimum Cost I (LeetCode #3010) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n) space complexity. Key insight: The cost of a subarray is determined solely by its first element.

What is the time complexity of Divide an Array Into Subarrays With Minimum Cost I?

The optimal solution for Divide an Array Into Subarrays With Minimum Cost I runs in O(n³) time and uses O(n) space. The time complexity is O(n³) due to the nested loops for filling the DP table, but this is significantly more efficient than the brute force approach.

What is the brute force solution for Divide an Array Into Subarrays With Minimum Cost I?

The brute force approach for Divide an Array Into Subarrays With Minimum Cost I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to split the array into three contiguous subarrays and calculating the total cost for each configuration. This method is straightforward but inefficient due to the large number of combinations. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Divide an Array Into Subarrays With Minimum Cost I?

Divide an Array Into Subarrays With Minimum Cost I uses the following concepts: Array, Sorting, Enumeration. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Divide an Array Into Subarrays With Minimum Cost I?

Always clarify the problem constraints and requirements. Think about edge cases and how they might affect your solution. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Divide an Array Into Subarrays With Minimum Cost I?

Not considering the boundaries when splitting the array. Overlooking the fact that the cost is based on the first element.

#3010

Divide an Array Into Subarrays With Minimum Cost I

Easy

ArraySortingEnumerationDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n)

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Intuition

Time O(n³)Space O(n)

The optimal solution uses dynamic programming to efficiently calculate the minimum cost by storing intermediate results. This avoids redundant calculations and significantly reduces the number of operations needed.

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Algorithm

3 steps

1Step 1: Create a DP array to store the minimum cost for splitting the array into k subarrays.
2Step 2: Iterate through the array and update the DP array based on previous results and the current element's cost.
3Step 3: Return the minimum cost for 3 subarrays.

solution.py10 lines

1def minCost(nums):
2    n = len(nums)
3    dp = [[float('inf')] * 4 for _ in range(n)]
4    for i in range(n):
5        dp[i][1] = nums[0] if i == 0 else min(dp[i-1][1], nums[0])
6    for k in range(2, 4):
7        for i in range(n):
8            for j in range(i):
9                dp[i][k] = min(dp[i][k], dp[j][k-1] + nums[i])
10    return dp[n-1][3]

ℹ

Complexity note: The time complexity is O(n³) due to the nested loops for filling the DP table, but this is significantly more efficient than the brute force approach.

1The cost of a subarray is determined solely by its first element.
2Dynamic programming can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.